Solving a First ODE Using an Integrating Factor

Turion · Sep 11, 2013

$$w'+2w=0\\ \frac { dw }{ dx } =-2w\\ I(x)={ e }^{ 2x }\\ \frac { dw }{ dx } { e }^{ 2x }=-2w{ e }^{ 2x }\\ \int { \frac { dw }{ dx } { e }^{ 2x } } dx=\int { -2w{ e }^{ 2x } } dx$$

Not sure what to do next.

Mark44 · Sep 11, 2013

Turion said: ↑

$$w'+2w=0\\ \frac { dw }{ dx } =-2w\\ I(x)={ e }^{ 2x }\\ \frac { dw }{ dx } { e }^{ 2x }=-2w{ e }^{ 2x }\\ \int { \frac { dw }{ dx } { e }^{ 2x } } dx=\int { -2w{ e }^{ 2x } } dx$$

Not sure what to do next.

Starting from dw/dx = -2w, you can use the method of separation, getting w and dw on one side, and x and dx on the other. You don't need an integration factor, since the equation is already separable.

BTW, the homework template is there for a reason. Don't just delete its parts. Fair warning...

Zondrina · Sep 11, 2013

You should get :

##w' + 2w = 0##
##w'e^{2x} + 2e^{2x}w = 0##

Notice the left hand side of the equation can be written :

##\frac{d}{dx} [e^{2x}w] = 0##

Try integrating now :)

Turion · Sep 11, 2013

This first-order DE is actually the result of a reduction of a second-order DE:

$$y''-y=0\quad \quad \quad \quad \quad { y }_{ 1 }={ e }^{ x }\\ Let\quad { y }_{ 2 }=u(x){ e }^{ x }\\ Sub\quad { y }_{ 2 },\quad { y }_{ 2 }',\quad and\quad { y }_{ 2 }''\quad into\quad the\quad DE\\ { e }^{ x }{ u }''+2{ e }^{ x }u'=0\\ Let\quad w=u'\\ w'+2w=0\\ dw=-2wdx\\ \frac { -1 }{ 2w } dw=dx\\ \int { \frac { -1 }{ 2w } dw } =\int { dx } \\ \frac { -1 }{ 2 } \int { \frac { -1 }{ w } dw } =\int { dx } \\ Ignoring\quad c\quad for\quad now\\ \frac { -1 }{ 2 } ln|w|=x\\ |w|={ e }^{ -2x }\\ |\frac { du }{ dx } |={ e }^{ -2x }$$

I was wondering what I should do to proceed?

Zondrina · Sep 11, 2013

Turion said: ↑

This first-order DE is actually the result of a reduction of a second-order DE:

$$y''-y=0\quad \quad \quad \quad \quad { y }_{ 1 }={ e }^{ x }\\ Let\quad { y }_{ 2 }=u(x){ e }^{ x }\\ Sub\quad { y }_{ 2 },\quad { y }_{ 2 }',\quad and\quad { y }_{ 2 }''\quad into\quad the\quad DE\\ { e }^{ x }{ u }''+2{ e }^{ x }u'=0\\ Let\quad w=u'\\ w'+2w=0\\ dw=-2wdx\\ \frac { -1 }{ 2w } dw=dx\\ \int { \frac { -1 }{ 2w } dw } =\int { dx } \\ ***\frac { -1 }{ 2 } \int { \frac { -1 }{ w } dw } =\int { dx }*** \\ Ignoring\quad c\quad for\quad now\\ \frac { -1 }{ 2 } ln|w|=x\\ |w|={ e }^{ -2x }\\ |\frac { du }{ dx } |={ e }^{ -2x }$$

I was wondering what I should do to proceed?

I see you took the separating variables route. You made a mistake where I highlighted with ***, there are two negatives when there is only one : ##\frac { -1 }{ 2 } \int { \frac { -1 }{ w } dw } =\int { dx }##

Also, you don't need the absolute value on the logarithm in the line below it.

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Solving a First ODE Using an Integrating Factor

Turion

Mark44

Staff: Mentor

Zondrina

Turion

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